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AVprozaik [17]
2 years ago
13

A small hole in the wing of a space shuttle requires a 20.7-cm² patch.

Chemistry
1 answer:
cricket20 [7]2 years ago
8 0

The Patch's area of the space shuttle in km² is 2.07 × 10⁻⁹ km²

Given, that a space shuttle requires a 20.7 cm² patch

We have to convert the patch's area from cm² into km².

Unit conversion is a method in which we multiply or divide with a particular numerical factor and then finally round off to the nearest significant digits.

Patch area of the space shuttle is 20.7 cm²

1 cm = 0.00001 km

or, 1 cm² = (0.00001 km)²

or,  1 cm² = 10⁻¹⁰km²

20.7 cm² = 20.7 ×  10⁻¹⁰km²

20.7 cm² = 2.07 × 10⁻⁹ km²

The patch area in square kilometers is 2.07 × 10⁻⁹ km²

To learn more about unit conversion, visit: brainly.com/question/11543684

#SPJ4

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Percentage by volume of solution is the percentage volume of solute in total volume of solution.
Volume percentage (v/v%) = volume of solute / total volume of solution x 100%
volume of solute - 16.0 mL
total volume of solution - 155 mL 
v/v% = 16.0 / 155 x 100% = 10.32%
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Explanation:

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Alex777 [14]

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S = 1.1 × 10⁻⁹ M

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We can find the molar solubility (S) of AgCl using an ICE chart.

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I                         0            0.15

C                      +S              +S

E                        S            0.15+S

The solubility product (Ksp) is:

Ksp = 1.6 × 10⁻¹⁰ = [Ag⁺].[Cl⁻] = S (0.15 + S)

If we solve the quadratic equation, the positive result is S = 1.1 × 10⁻⁹ M

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